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Let $A \colon= K[[X_1,\ldots,X_m,Y_1,\ldots,Y_n]]$ be a power series ring over a field $K$ in $m + n$ variables and ${\frak m}$ be the unique maximal ideal of $A$.

For arbitrary two elements $\alpha = X_1^{e_1} \cdots X_m^{e_m}$, $\beta = Y_1^{f_1} \cdots Y_n^{f_n}$ with $e_i, f_j \geq 1$, we shall add $a \in {\frak m}$ to $\alpha$ and $b \in {\frak m}$ to $\beta$ such that $a$ (resp. $b$) comprises scalar times monomials different from $\alpha$ (resp. $\beta$). Consequently, $a + \alpha \in {\frak m}$ (resp. $b + \beta \in {\frak m}$) contains $\alpha$ (resp. $\beta$) as its non-trivial constituent.

Q. Does the following sufficiently large constant $C \gg 0$ which is independent of $m, n$ exist? :

The product $(a + \alpha)(b + \beta)$ always involves a nonzero monomial $X_1^{v_1} \cdots X_m^{v_m}Y_1^{w_1} \cdots Y_n^{w_n}$ (up to scalar coefficient) such that inequalities $v_i \leq Ce_i$, $w_j \leq Cf_j$ for $1 \leq i \leq m$ and $1 \leq j \leq n$ hold simultaneously.

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Yes if you allow $C$ to depend on $n$ and $m$. Define the graded degree of a monomial$\prod X_i^{v_i}\prod Y_j^{w_j}$ as $\sum v_i/e_i+\sum w_j/f_j$. Then the first bracket $a+\alpha$ contains the monomial $\alpha$ with graded degree at most $m$, the second bracket $b+\beta$ contains the monomial $\beta$ with graded degree at most $n$, thus the product $(a+\alpha)(b+\beta)$ contains a monomial with graded degree at most $m+n$ (the products of monomials of minimal graded degree do not cancel.) This proves your claim for $C=m+n$.

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  • $\begingroup$ Great thanks! But I wish C to be independent of $m, n$. Rinmyaku $\endgroup$ – Rinmyaku Jul 13 at 10:10
  • $\begingroup$ Dear Professor Fedor Petrov, thanks. It suffices to get your result. Pierre Matsumi $\endgroup$ – Rinmyaku Jul 14 at 9:35

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